Question

A particle on a stretched string supporting a travelling wave, takes minimum time $5.0ms$ to move from its mean position to the mean position. The distance between two consecutive particles, which are at their mean position, is $2.0cm$. Find the frequency, the wavelength and the wave speed.

Answer 1

As the particle two consecutive mean position in $0.5ms$,

$\Rightarrow \frac{T}{2}=0.5ms\Rightarrow T=1ms$

So, frequency. $V=\frac{1}{T}=\frac{1}{{10}^{-3}}={10}^{3}HZ$

Two consecutive mean position are located $2.0cm$ apart,

$\Rightarrow \frac{\lambda }{2}=2cm\Rightarrow \lambda =4cm$

So wave velocity, $v=\lambda v=4\times {10}^{-2}\times {10}^3$

$\Rightarrow V=40m/s.$